{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nPotentiometer method of DC voltage measurement is more accurate than direct measurement using a voltmeter because\n(A) It loads the circuit moderately. (B) It loads the circuit to maximum extent. (C) It uses centre zero galvanometer instead of voltmeter. (D) It does not load the circuit at all. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of these sets of logic gates are designated as universal gates?\n(A) NOR, NAND. (B) XOR, NOR, NAND. (C) OR, NOT, AND. (D) NOR, NAND, XNOR. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA single phase one pulse controlled circuit has a resistance R and counter emf E load 400 sin(314 t) as the source voltage. For a load counter emf of 200 V, the range of firing angle control is\n(A) 30\u00b0 to 150\u00b0. (B) 30\u00b0 to 180\u00b0. (C) 60\u00b0 to 120\u00b0. (D) 60\u00b0 to 180\u00b0. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA box which tells the effect of inputs on control sub system is known as\n(A) Data Box. (B) Logical box. (C) Decision box. (D) State box. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nTwo infinite parallel metal plates are charged with equal surface charge density of the same polarity. The electric field in the gap between the plates is\n(A) same as that produced by one plate. (B) double the field produced by one plate. (C) dependent on coordinates of field points. (D) zero. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe resistance in the circuit of the moving coil of a dynamometer wattmeter should be\n(A) Low. (B) Very low. (C) High. (D) Almost zero. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe following are the necessary requirements of an oscillator\n(A) Amplitude stability (B) Frequency stability (C) Power stability (D) Both A and B \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe normal voltage used in directional heating is\n(A) 1.5 KV. (B) 15 KV. (C) 33 KV. (D) 66 KV. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nLC oscillators are used for produce a waveform with frequency ranging from\n(A) 1MHz to 500 MHz (B) 100 KHz to 500 MHz (C) 1 KHz to 1 MHz (D) 1MHz to 100 GHz \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA working diode must have\n(A) High resistance when forward or reverse biased (B) Low resistance when forward biased, while high resistance when reverse bias (C) High resistance when forward biased, while low resistance when reverse bias (D) Low resistance when forward or reverse biased \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhat is the name of the fluorescent material that gives red colour fluorescence?\n(A) Zinc silicate. (B) Calcium silicate. (C) Zinc sulphide. (D) Magnesium silicate. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nHow many bits are required to store one BCD digit ?\n(A) 1 (B) 2 (C) 3 (D) 4 \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nCommutator in DC generator is used for\n(A) collecting of current (B) reduce losses (C) increase efficiency (D) convert AC armature current in to DC \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\n1 F is theoretically equal to\n(A) 1 ohm of resistance (B) ratio of 1 V to 1 C (C) ratio of 1 C to 1 V (D) none of these \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nHow many entries will be in the truth table of a 3 input NAND gate ?\n(A) 3 (B) 6 (C) 8 (D) 9 \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn decimal number system what is MSD\n(A) First digit from left to right (B) First digit from right to left (C) Middle digit (D) Mean of all digits \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn Digital Logic Designs, GAL is abbreviated as\n(A) General Advance Logic (B) General Array Logic (C) Generic Advance Logic (D) Generic Array Logic \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nInintel 8085A microprocessor ALE signal is made high to\n(A) Enable the data bus to be used as low order address bus (B) To latch data D0-D7 from data bus (C) To disable data bus (D) To achieve all the functions listed above \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nFor an SCR gate cathode characteristic is a straight line of 130. For triggered source volume of 15 V and allowable gate power dissipation of 0.5 W compute the gate source resistance?\n(A) 111.9 ohm (B) 11.19 ohm (C) 108 ohm (D) 115 ohm \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe concept of V/f control of inverters driving induction motors resuls in\n(A) constant torque operation (B) speed reversal (C) reduced magnetic loss (D) hormonic elimination \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe energy stored in the magnetic field in a solenoid 30 cm long and 3 cm diameter wound with 1000 turns of wire carrying a current at 10 amp, is\n(A) 0.015 joule. (B) 0.15 joule. (C) 0.015 joule. (D) 1.15 joule. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhile operating on variable frequency supplies, the AC motor requires variable voltage in order to\n(A) protect the insulation. (B) avoid effect of saturation. (C) improve the capabilities of the inverter. (D) protect thyristor from dV / dt. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following windings are necessary in case of all dc machines?\n(A) closed winding (B) lap winding (C) wave winding (D) open type winding \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nRelaxation Voltage controlled oscillators are used to generate\n(A) Sinusoidal wave (B) Triangular wave (C) Sawtooth wave (D) Both B and C \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe supply frequency usually employed for high frequency eddy current heating is\n(A) 1 KHz. (B) 5 KHz. (C) 10 MHz. (D) 10 KHz to 400 KHz. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn the toggle mode a JK flip-flop has\n(A) J = 0, K = 0. (B) J = 1, K = 1. (C) J = 0, K = 1. (D) J = 1, K = 0. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nElectric drive is becoming more and more popular because\n(A) all of below. (B) it provide smooth and easy control. (C) it is cheaper in cost. (D) it is simple and reliable. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA step up chopper has input voltage 110 V and output voltage 150 V. The value of duty cycle is\n(A) 0.32. (B) 0.67. (C) 0.45. (D) 0.27. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA Binary number system has how many digits.\n(A) 0 (B) 1 (C) 2 (D) 10 \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nLowest critical frequency is due to pole and it may be present at the origin or nearer to the origin, then the type of network is\n(A) LC. (B) RL. (C) RC. (D) Any of the above. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nAdvantages of open loop control system is/are\n(A) simple and economical. (B) accurate. (C) reliable. (D) all of the above. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe voltage induced in an inductor is represented as,\n(A) product of its inductance and current through it. (B) ratio of its inductance to current through it. (C) ratio of current through it to its inductance. (D) product of its inductance and rate of change of current through it. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIf a Hexadecimal number needs to convert to binary. For each hexadecimal digit, there will be how many bits\n(A) 1 (B) 2 (C) 4 (D) 8 \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nInduction heating takes place in\n(A) conducting but non-magnetic materials. (B) conducting materials which may be either magnetic or non-magentic materials. (C) insulating materials. (D) conducting and magnetic materials. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn a lap winding dc machine number of conductors are 100 and number of parallel paths are 10. Find the average pitch\n(A) 10 (B) 100 (C) 50 (D) 1 \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nMagnetic flux has the unit of\n(A) Newton (B) Ampere turn (C) Weber (D) Tesla \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn balanced bridge, if the positions of detector and source are interchanged, the bridge will still remain balanced. This can be explained from which theorem\n(A) Reciprocity theorem (B) Thevinin's theorem (C) Norton's theorem (D) Compensation theorem \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA D-flip-flop is said to be transparent when\n(A) the output is LOW (B) the output is HIGH (C) the output follows clock (D) the output follow input \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA Lissajous pattern on an oscilloscope has 5 horinzontal tangencies and 2 vertical tangencies. The frequency of horizontal input is 100 Hz. The frequency of the vertical will be\n(A) 500 Hz. (B) 250 Hz. (C) 400 Hz. (D) 625 Hz. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nSpeed control by variation of field flux results in\n(A) constant power drive. (B) constant torque drive. (C) variable power drive. (D) none of the above. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following memories uses one transistor and one capacitor as basic memory unit\n(A) SRAM (B) DRAM (C) Both SRAM and DRAM (D) nan \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhat will be the number of lamps, each having 300 lumens, required to obtain an average illuminance of 50 lux on a 4m * 3m rectangular room?\n(A) 2 (B) 4 (C) 5 (D) 6 \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe output of the system has an effect upon the input quantity, then the system is a\n(A) open loop control system. (B) closed loop control system. (C) either A or B. (D) none of the above. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA sinusoidal signal is analog signal, because\n(A) it can have a number of values between the negative and positive peaks (B) it is negative for one half cycle (C) it is positive for one half cycle (D) it has positive as well as negative values \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nFM stands for\n(A) frequent modulation. (B) frequency modulation. (C) frequency moderator. (D) frequent moderator. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nStack is also known as\n(A) FIFO memory (B) Flash memory (C) LIFO memory (D) LILO memory \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn Ward-Leonard system, the lower limit of the speed imposed by\n(A) Field resistance. (B) Armature resistance. (C) Residual magnetism of the generator. (D) None of above. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nColpitts oscillator is also called as\n(A) Tank circuit oscillator (B) LC oscillator (C) Resonant circuit oscillator (D) All of the above \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nAn SCR has half cycle surge current rating of 3000 A for 50 Hz supply. One cycle surge current will be\n(A) 1500 A. (B) 6000 A. (C) 2121.32 A. (D) 4242.64 A. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nBoth the ALU and control section of CPU employ which special purpose storage location?\n(A) Buffers (B) Decoders (C) Accumulators (D) Registers \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nSystematic errors are\n(A) environmental errors. (B) observational errors. (C) instrument errors. (D) all of the above. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following is/are the disadvantages of an open loop control system?\n(A) Inaccurate. (B) Unreliable. (C) both A and B. (D) none of the above. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA diode (p-n junction) when reverse biased act as\n(A) On Switch (B) Zener diode (C) Capacitor (D) Off Switch \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nPiezoelectric effect is carried out in\n(A) composite filter. (B) crystal filter. (C) m derived. (D) constant k prototype. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe resistivity of the conductor depends on\n(A) area of the conductor. (B) length of the conductor. (C) type of material. (D) none of these. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe smallest change in measured variable to which instrument will respond is\n(A) resolution. (B) accuracy. (C) precision. (D) sensitivity. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn electronic communications, \"AM\" stands for\n(A) Amp Modification (B) Amplitude Method (C) Amplitude Modulation (D) Ampere Method \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich system is also known as automatic control system?\n(A) Open loop control system (B) Closed loop control system (C) Either A or B (D) Nether A nor B \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA combinational PLD with a fixed AND array and a programmable OR array is called a\n(A) PLD (B) PROM (C) PAL (D) PLA \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA null type instrument as compared to a deflected type instrument has\n(A) a lower sensitivity. (B) a faster response. (C) a higher accuracy. (D) all of the above. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nLength of the cable is doubled, its capacitance C will be\n(A) one-fourth. (B) one-half. (C) doubled. (D) unchanged. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nElectrons in p-type material of a semi-conductor are called as\n(A) either minority carriers or majority carriers (B) minority carriers (C) majority carriers (D) valance carriers \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\n________ is the most detrimental impurity in the magnetic materials\n(A) Carbon. (B) Sulphur. (C) Oxygen. (D) Nitrogen. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nVoltage controlled oscillators are used commonly in\n(A) Pulse Modulators Frequency Modulators Phase Clocked loops (B) Frequency Modulators (C) Phase Clocked loops (D) All the above \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nLow frequency supply is necessary for direct core type induction furnaces because\n(A) With normal frequeny supply the electromagnetic forces causes severe stirring action in the molten metal. (B) Magnetic coupling between the primary and secondary circuit is poor. (C) Both A and B. (D) None of the above. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe moving coil-meters, damping is provided by\n(A) the aluminium frame of the coil. (B) the coil spring attached to the moving. (C) eddy current disk. (D) damping vane in the airtight chamber. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following is the analogous quantity for mass element in force-voltage analogy?\n(A) Resistance. (B) Inductance. (C) Capacitance. (D) All of the above. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA latch is ________ sensitive\n(A) both level and edge (B) edge (C) level (D) nan \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhat is used to protect the SCR from over current ?\n(A) CB and fuse. (B) Heat sink. (C) Snubber circuit. (D) Voltage clamping device. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe errors mainly caused by human mistakes are\n(A) gross error. (B) instrumental error. (C) observational error. (D) systematic error. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nElectrical analogous quantity for dash-pot in force-current analogy is\n(A) resistance. (B) conductance. (C) inductance. (D) capacitance. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich lamp has the best Colour Rendering Index (CRI)?\n(A) LED (B) Fluorescent (C) Incandescent (D) High pressure sodium vapour \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nPower dissipation in ideal inductor is\n(A) Maximum (B) Minimum (C) Zero (D) A finite value \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nFor intermittent work which of the following furnace is suitable?\n(A) Core less furnace. (B) Indirect arc furnace. (C) Either of above. (D) Nither of above. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn electrical resistance welding material of electrode should have\n(A) higher electrical conductivities. (B) higher thermal conductivities. (C) sufficient strength to sustain high pressure at elevated temperatures. (D) all of above. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following are the passive elements?\n(A) Resistor (B) Bulb (C) Both (D) None of these. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe main advantage of dielectric heating is that\n(A) heating occurs in the material itself. (B) heating occurs due to high frequency. (C) it can be used for drying the explosives. (D) None of the above. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nMaterials used in heat sink should have\n(A) high thermal conductivity. (B) large surface area. (C) high melting point. (D) All of these. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nConductor is static and the field is varying then emf will be induced. This principle is called\n(A) virtually induced emf. (B) dynamically induced emf. (C) static induced emf. (D) none of these \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nFour identical alternators each are rated for 20 MVA, 11 KV having a subtransient reactance of 16% are working in parallel. The short circuit level at the busbar is\n(A) 500 MVA (B) 400 MVA (C) 125 MVA (D) 100 MVA \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nAccording to the Bohr model, an electron gains or losses energy only by\n(A) moving faster or slower in an allowed orbit. (B) jumping from one allowed orbit to another. (C) being completely removed from an atom. (D) jumping from one atom to another atom. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA three-state buffer has the following output states\n(A) 1, 0, float (B) High, Low, Float (C) Both A and B (D) Set, Reset, Halt \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nBy using which of the following elements, mechanical translational systems are obtained?\n(A) mass element (B) spring element (C) dash-pot (D) all of the above \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nFour capacitors each of 40 \u00b5F are connected in parallel, the equivalent capacitance of the system will be\n(A) 160 \u00b5F (B) 10 \u00b5F (C) 40 \u00b5F (D) 5 \u00b5F \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe resistance of a conductor of diameter d and length l is R \u03a9. If the diameter of the conductor is halved and its length is doubled, the resistance will be\n(A) R \u03a9 (B) 2R \u03a9 (C) 4R \u03a9 (D) 8R \u03a9 \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA single phase full bridge inverter can operated in load commutation mode in case load consist of\n(A) RL. (B) RLC underdamped. (C) RLC overdamped. (D) RLC critically damped. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA microprocessor is ALU\n(A) and control unit on a single chip. (B) and memory on a single chip. (C) register unit and I/O device on a single chip. (D) register unit and control unit on a single chip. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nTo obtain a high value of capacitance, the permittivity of dielectric medium should be\n(A) low (B) zero (C) high (D) unity \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nDeflection method direct measurements are most widely used as these are\n(A) least time consuming. (B) most simple. (C) most simple and least time consuming. (D) most accurate. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nPurely mechanical instrument cannot be used for dynamic measurements because they have\n(A) high inertia. (B) higher response time. (C) large time constant. (D) all of the above. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nAdvantages of higher transmission voltage is/are\n(A) Power transfer capability of the transmission line is increased (B) Transmission line losses are reduced (C) Area of cross section and volume of the conductor is reduced (D) all of the above \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nUnder over voltage condition impedance offered by the voltage clamping device is\n(A) high. (B) low. (C) moderate. (D) infinity. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA long wire composed of a smooth round conductor runs above and parallel to the ground (assumed to be a large conducting plane). A high voltage exists between the conductor and the ground. The maximum electric stress occurs at\n(A) lower surface of the conductor. (B) upper surface of the conductor. (C) the ground surface. (D) midway between the conductor and ground. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn a dc machine 6 pole wave winding is used. The number of parallel paths are?\n(A) 6 (B) 4 (C) 2 (D) 1 \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following is a digital device\n(A) Regulator of a fan (B) Microphone (C) Resistance of a material (D) Light switch \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nInstantaneous power in inductor is proportional to the\n(A) product of the instantaneous current and rate of change of current. (B) square of instantaneous current. (C) square of the rate of change of current. (D) temperature of the inductor. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following is a variable displacement transducer?\n(A) tachometer (B) potentiometer (C) synchros (D) none of the above \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nFive capacitors each of 5 \u00b5F are connected in series, the equivalent capacitance of the system will be\n(A) 5 \u00b5F (B) 25 \u00b5F (C) 10 \u00b5F (D) 1 \u00b5F \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nSusceptibility is negative for\n(A) non magnetic substances. (B) diamagnetic substances. (C) ferromagnetic substances. (D) none of above. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nOne of the following is the primary function of an oscillator\n(A) produces sinusoidal oscillations (B) generates non sinusoidal waveforms (C) generates sustained oscillations at a constant amplitude and specific frequency (D) none of the above \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe metal surfaces for electrical resistance welding must be\n(A) rough. (B) clean. (C) moistened (D) coloured \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe latching current of SCR is 20 mA. Its holding current will be\n(A) 23 mA. (B) 40 mA. (C) 10 mA. (D) 60 mA. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn force-current analogy, electrical analogous quantity for displacement (x) is\n(A) voltage. (B) inductance. (C) capacitance. (D) flux. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nFeedback circuit in an oscillator can be accomplished by\n(A) Resistive coupling between input and output (B) Capacitive coupling between input and output (C) Inductive coupling between input and output (D) Any one or combinations of the above methods \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nLowest critical frequency is due to zero and it may be present at the origin or nearer to the origin, then the type of network is\n(A) LC circuit. (B) RC circuit. (C) RLC circuit. (D) RL Circuit. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhat is the value of dielectric constant of air?\n(A) Less than 1 (B) 0. (C) 1. (D) none of these. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nTemperature variation is a/an\n(A) Digital quantity (B) Analog quantity (C) Either Digital or Analog quantity (D) nan \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe illumination at a point 5 meters below a lamp in 6 lux. The candle power of the lamp is\n(A) 30 (B) 140. (C) 150. (D) 200. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nFor a dc machine shunt resistance and armature resistance values are\n(A) high and high (B) high and low (C) low and low (D) low and high \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe change of cross sectional area of conductor in magnetic field will affect\n(A) reluctance of conductor. (B) resistance of conductor. (C) (A) and (B) both in the same way. (D) none of above. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following is used in SCR to protect from high dV / dt?\n(A) Snubber circuit. (B) Fuse. (C) Equalizing circuit. (D) Circuit breaker. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhat are the sets of commands in a program which are not translated into machine instructions during assembly process, called?\n(A) Mnemonics (B) Directives (C) Identifiers (D) Operands \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nSpeed of data transmission in 4-G network of telecom is\n(A) 386 kbps - 2 mbps. (B) 2 mbps. (C) 2 mbps \u2013 1 gbps. (D) 100 mbps - 1 gbps. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the followings is/are active element?\n(A) Voltage source (B) Current source (C) Both (D) None of these. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nElectrical analogous quantity for spring element (K) in force-voltage analogy is\n(A) L (B) R (C) 1/C (D) C \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA quartz crystal oscillator consists of\n(A) only series resonant frequency. (B) only parallel resonant frequency. (C) both series and parallel frequencies. (D) neither series nor parallel frequency. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following oscillators are used for low frequency (LF) applications\n(A) LC oscillators (B) RC oscillators (C) Both LC and RC Oscillators (D) nan \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe solid angle subtended by an area of 2400 cm^2 on the surface of a sphere of diameter 1.2 m is\n(A) 3 / 2. (B) 1 / 3. (C) 2 / 3. (D) 2 / 5. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nRelative error is same as\n(A) ratio of absolute error and true value. (B) absolute error. (C) true error. (D) none of the above. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIf holding current of a thyristor is 2 mA then latching current should be\n(A) 0.01 A. (B) 0.002 A. (C) 0.009 A. (D) 0.004 A. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn an intel 8085A, which is the first machine cycle of an instruction?\n(A) An op-code fetch cycle (B) A memory read cycle (C) A memory write cycle (D) An I/O read cycle \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following are the disadvantages of a closed loop control system?\n(A) Reduces the overall gain. (B) Complex and costly. (C) Oscillatory response. (D) All of the above. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nPoles and zeros are arranged alternatively on negative real axis, then type of network is/are\n(A) LC network. (B) RC network. (C) RL network. (D) Both 2 and 3. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe magnetisation and applied field in ferromagnetic materials are related\n(A) sinusoidally. (B) non linearly. (C) linearly. (D) parabolically. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThin laminations are used in a machine in order to reduce\n(A) Eddy current losses (B) Hysteresis losses (C) Both A and B (D) Copper losses \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA metal surface with 1 meter radius and surface charge density of 20 coulombs / m2 is enclosed in a 10 m side cube. The total outward electric displacement normal to the surface of the cube is\n(A) 40\u03c0 coulombs. (B) 80\u03c0 coulombs. (C) 10\u03c0 coulombs. (D) none of these. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn binary number system the first digit (bit) from right to left is called as\n(A) LSB, Least Significant Bit (B) MSB, Most Significant Bit (C) First Bit (D) Last Bit \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn two wattmeter method of power measurement if the total power is measured by one wattmeter only then power factor of the system is\n(A) 0. (B) 0.5. (C) 1. (D) none of above. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA 0 - 100 V voltmeter has a guaranteed accuracy of 2 % of full scale reading. The voltage measured by the voltmeter is 75 V. The limiting error is in percentage\n(A) 3.33 %. (B) 2.66 %. (C) 2 %. (D) 1 %. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following methods is/are used for reactive or voltage compensation\n(A) shunt capacitor (B) series capacitor (C) generation excitation control (D) all of the above \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe ROM programmed during manufacturing process itself is called\n(A) MROM (B) PROM (C) EPROM (D) EEPROM \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIn 8085 name/names of the 16 bit registers is/are\n(A) stack pointer. (B) program counter. (C) both A and B. (D) none of these. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nStability of a transmission line can be increased by\n(A) shunt capacitor (B) series capacitor (C) shunt reactor (D) both A and B \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nSnubber circuit is used with SCR\n(A) in series. (B) in parallel. (C) either series or parallel. (D) anti parallel. \nA: Let's think step by step.", "label": "(B)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe number of rings in the Bohr model of any element is determined by what?\n(A) Column number on periodic table. (B) Atomic mass. (C) Row number on periodic table. (D) Atomic number. \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nSilicon and Germanium are ________ elements.\n(A) trivalant (B) pentavalant (C) hexavalant (D) tetravalant \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe number of output pins in 8085 microprocessors are\n(A) 27. (B) 40. (C) 21. (D) 19. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIf P is the power of a star connected system then what will be power of an equivalent delta connected system?\n(A) P (B) 3P (C) P/3 (D) None of the above \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich oscillators are easy to fabricate in a monolithic IC?\n(A) Crystal oscillator. (B) Hartley oscillator. (C) Wien bridge oscillator. (D) Relaxation oscillator. \nA: Let's think step by step.", "label": "(D)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nIf all the elements in a particular network are linear, then the superposition theorem would hold, when the excitation is\n(A) DC only (B) AC only (C) Either AC or DC (D) An Impulse \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe feedback factor of a Wien bridge oscillator using Op-Amp is\n(A) 1/3 (B) 1/4 (C) 1/2 (D) 1 \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nThe length of bus cycle in 8086/8088 is four clock cycles, T1, T2, T3, T4 and an indeterminate number of wait state clock cycles denoted by Tw. The wait states are always inserted between\n(A) T1 & T2 (B) T2 & T3 (C) T3 & T4 (D) T4 & T1 \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nWhich of the following is not a sinusoidal oscillator?\n(A) LC oscillator (B) RC phase shift oscillator (C) Relaxation oscillator (D) Crystal oscillator \nA: Let's think step by step.", "label": "(C)", "options": ["(A)", "(B)", "(C)", "(D)"]}
{"question": "\nThe following are multiple choice questions (with answers) about electrical engineering.\n\nQ: A point pole has a strength of 4\u03c0 * 10^-4 weber. The force in newtons on a point pole of 4\u03c0 * 1.5 * 10^-4 weber placed at a distance of 10 cm from it will be\n(A) 15 N. (B) 20 N. (C) 7.5 N. (D) 3.75 N.\nA: Let's think step by step. The force between two point poles is given by m_1m_2/(mu_0 4 \\pi r^2), in analogy to Coulomb\u2019s law. Plugging in the values given in the question, we calculate that the force is approximately 15 N. The answer is (A).\n\nQ: The coil of a moving coil meter has 100 turns, is 40 mm long and 30 mm wide. The control torque is 240*10-6 N-m on full scale. If magnetic flux density is 1Wb/m2 range of meter is\n(A) 1 mA. (B) 2 mA. (C) 3 mA. (D) 4 mA.\nA: Let's think step by step. The torque on a coil in a uniform magnetic field is given by BANI, where B is the magnetic flux density, A is the area of the coil, N is the number of turns, and I is the current. So we have that I = (Torque)/(BAN), or 240e-6/(1200e-6 * 100 * 1) = 2e-3. The answer is (B).\n\nQ: In an SR latch built from NOR gates, which condition is not allowed\n(A) S=0, R=0 (B) S=0, R=1 (C) S=1, R=0 (D) S=1, R=1\nA: Let's think step by step. An SR latch is a set-reset latch; in the case where S=1 and R=1, the circuit has no stable state; instead a race condition will be produced within the circuit, so the device will be in an undefined state. So S=1, R=1 is an illegal input. The answer is (D).\n\nQ: Two long parallel conductors carry 100 A. If the conductors are separated by 20 mm, the force per meter of length of each conductor will be\n(A) 100 N. (B) 0.1 N. (C) 1 N. (D) 0.01 N.\nA: Let's think step by step. The magnetic force-per-length between two current-carrying conductors is given by \\mu_0 I_1 I_2 / (2 \\pi r), where $r$ is the separation distance and I_1 and I_2 are the currents. Plugging in 100 A for I_1 and I_2, and 20 mm for r, gives 0.1 N. The answer is (B).\n\nQ: In a 2 pole lap winding dc machine , the resistance of one conductor is 2\u03a9 and total number of conductors is 100. Find the total resistance\n(A) 200\u03a9 (B) 100\u03a9 (C) 50\u03a9 (D) 10\u03a9\nA: Let's think step by step. In lap winding, effectively two resistors are connected in parallel, so the actual resistance of each pair is 1 Ohm. Since we have 50 pairs, we get a total resistance of 50 Ohms. The answer is (C).\n\nA wire carrying current is bent in the form of a circular loop. Then the magnetic field around each portion of the wire will be\n(A) parallel to the plane of the wire. (B) perpendicular to the circumference of the wire. (C) parallel to half portion and perpendicular for the other half. (D) none of above. \nA: Let's think step by step.", "label": "(A)", "options": ["(A)", "(B)", "(C)", "(D)"]}
